Electric-Light
Senior Member
v = 200v rms
i = 5A rms
no further information given
i = 5A rms
no further information given
What is the power of an incandescent lamp at 200v 5A?
- Thread starter Electric-Light
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gar
Senior Member
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This appears to be a trick question based on the strange wording of the question, and that the quick obvious answer is the first in the answer list. Based on the possibility of a trick question the third answer is probably a better answer.
The question could be what is the approximate power input at 60 Hz, or the visible spectrum power output.
Then it might be what is the power input at a frequency of 1000 megacycles/sec. At 30 megacycles/sec the power input is probably fairly close to 1 KW. But not too close a 1 gHz. Certainly the inductance of the bulb becomes significant at high frequencies. But incandescent bulbs have been used as a dummy loads for RF transmitters to estimate power output, and other purposes, even into the 100 megaHz range.
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This appears to be a trick question based on the strange wording of the question, and that the quick obvious answer is the first in the answer list. Based on the possibility of a trick question the third answer is probably a better answer.
The question could be what is the approximate power input at 60 Hz, or the visible spectrum power output.
Then it might be what is the power input at a frequency of 1000 megacycles/sec. At 30 megacycles/sec the power input is probably fairly close to 1 KW. But not too close a 1 gHz. Certainly the inductance of the bulb becomes significant at high frequencies. But incandescent bulbs have been used as a dummy loads for RF transmitters to estimate power output, and other purposes, even into the 100 megaHz range.
.
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If this is a trick question then I guess I fell for it.
gar
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My point is that the wording of the question and the answers make the intent not clear.
My illustration on frequency is that if you apply a sufficiently high frequency then the internal inductance of the bulb may become significant and now you have to include the unknown power factor to determine the input power to the bulb vs input VA.
nakulak:
Also if you operate at a sufficiently low frequency so that the bulb resistance is non-linear thru the cycle, then RMS volts * RMS current does not necessarily equal input power.
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My point is that the wording of the question and the answers make the intent not clear.
My illustration on frequency is that if you apply a sufficiently high frequency then the internal inductance of the bulb may become significant and now you have to include the unknown power factor to determine the input power to the bulb vs input VA.
nakulak:
Also if you operate at a sufficiently low frequency so that the bulb resistance is non-linear thru the cycle, then RMS volts * RMS current does not necessarily equal input power.
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1. In which case the OP can go soak his head.
2. It seemed to me to be too simple to answer unless it was some sorta trick
....see #1.
3. If the frequency were too low, would it even function as a suitable incandescent
lamp or a strobe light? ... see #1.
4. If you state your problem in terms that aren't suitable for the problem.... see #1.
At a basic ohms law level Dennis answered the question in the second post, and with no more information given, it is not a trick question as it is asked.
There is no need to take this to any other level unless the OP comes back with added wording.
With that said, feel free to continue with the over thinking discussion. :grin:
Roger
There is no need to take this to any other level unless the OP comes back with added wording.
With that said, feel free to continue with the over thinking discussion. :grin:
Roger
I thought that was the purpose of RMS .... so that you would be using comparable units to get comparable units answers...and that the plain ol' average of a wave that was symetrical about the zero line would be zero .....
Although I'm over my head here, I also think that if you express the wave form in appropriate "Calculus-esque" terms, as it changes over time, you can RMS it, and it'll come out right.... but don't ask me to do it.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Agreed. 1kw. Now, let's hear from the OP.
Overthinking??? In this forum??? Never. :roll:
mivey
Senior Member
It is a sickness. It really is. :grin:
gar
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nakulak:
RMS is an averaging technique and is related to a means to get common readings for power between some AC and pure DC circuits. But you can not just multiply Vrms across a device by Irms into the device and from this product get the power input to the device. Sometimes you can and sometimes you can't. You need enough information and an understanding of how to calculate power to determine whether the simple product produces a correct answer or not.
I doubt that I can make a light bulb draw the following current without an electronic circuit associated with the bulb. But this description will show that Vrms * Irms does not necessarily equal power. My use of the word power without modifiers means real power.
In the following example my source voltage is a pure 60 hZ sine wave. Period for one cycle is 16.67 seconds. The current to the load is a rectangular pulse 0.1667 seconds long. Assume the RMS voltage is 100 V, and thus the peak is sq-root of 2 * 100 = 1.414 V.
Suppose my load draws current at the voltage peaks of the input voltage, and during this time it is about 1 A. What is the approximate RMS value of this current? About 0.1 A. 100 V * 0.1 = 10. What is the approximate actual average power to the load? 141.4 * 1 * 0.1667 / 16.67 = 1.414 W. These are quite different.
Next move the current pulse to a 45 deg point, but keep the RMS current level the same. Vrms * Irms is still 10, but actual power is about 100 * 1 / 100 = 1 W.
Last go to 0 deg with the same current pulse. The result is almost 0 W.
The red flag relative to the original post is the strange wording of the question and the strange relationship between the question and the possible answers.
In another thread the poster answered a test question and his instructor marked his answer as wrong. The instructor was correct. The question of this other thread might have been considered a trick question, but I do not put it in that category. Rather it was a good question to determine whether the student really understood the difference between power and energy.
The question of this thread seems to be more like --- can I trick you into answering 1 KW.
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nakulak:
RMS is an averaging technique and is related to a means to get common readings for power between some AC and pure DC circuits. But you can not just multiply Vrms across a device by Irms into the device and from this product get the power input to the device. Sometimes you can and sometimes you can't. You need enough information and an understanding of how to calculate power to determine whether the simple product produces a correct answer or not.
I doubt that I can make a light bulb draw the following current without an electronic circuit associated with the bulb. But this description will show that Vrms * Irms does not necessarily equal power. My use of the word power without modifiers means real power.
In the following example my source voltage is a pure 60 hZ sine wave. Period for one cycle is 16.67 seconds. The current to the load is a rectangular pulse 0.1667 seconds long. Assume the RMS voltage is 100 V, and thus the peak is sq-root of 2 * 100 = 1.414 V.
Suppose my load draws current at the voltage peaks of the input voltage, and during this time it is about 1 A. What is the approximate RMS value of this current? About 0.1 A. 100 V * 0.1 = 10. What is the approximate actual average power to the load? 141.4 * 1 * 0.1667 / 16.67 = 1.414 W. These are quite different.
Next move the current pulse to a 45 deg point, but keep the RMS current level the same. Vrms * Irms is still 10, but actual power is about 100 * 1 / 100 = 1 W.
Last go to 0 deg with the same current pulse. The result is almost 0 W.
The red flag relative to the original post is the strange wording of the question and the strange relationship between the question and the possible answers.
In another thread the poster answered a test question and his instructor marked his answer as wrong. The instructor was correct. The question of this other thread might have been considered a trick question, but I do not put it in that category. Rather it was a good question to determine whether the student really understood the difference between power and energy.
The question of this thread seems to be more like --- can I trick you into answering 1 KW.
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skeshesh
Senior Member
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Gar,
The result of your analysis shows inconsistencies related to RMS versus average value. Although these matters of statistics do matter, i think the question in this example is answered more easily.
I think 1kw is indeed misleading, but only as far as leading overanalysis of the subject, as you did, to arrive at the wrong conclusion.
Don't get me wrong, I appreciate the merits of your discussion (not to mention the immense amount of knowledge i've gained about VFDs, control circuits, etc. involving you, jraef, zog, charlie, etc.) but I do think this specific example is one of futility and the answer is as simple as it seems.
The result of your analysis shows inconsistencies related to RMS versus average value. Although these matters of statistics do matter, i think the question in this example is answered more easily.
I think 1kw is indeed misleading, but only as far as leading overanalysis of the subject, as you did, to arrive at the wrong conclusion.
Don't get me wrong, I appreciate the merits of your discussion (not to mention the immense amount of knowledge i've gained about VFDs, control circuits, etc. involving you, jraef, zog, charlie, etc.) but I do think this specific example is one of futility and the answer is as simple as it seems.
gar
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skeshesh:
There is no inconsistency between an average measurement and an RMS measurement. They are each what they are. To make proper use of these measurements you need to know the circuit and what is the goal of the measurements you are making. You get non-equal results in some equations when you use one method of measurement vs the other, and sometimes the same result occurs.
The question of this thread looks very much like it was a question on a test and Electric-Light wanted to see how members of this forum would respond.
What is your answer to the following question?
What is the maximum phase shift of an RC network?
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skeshesh:
There is no inconsistency between an average measurement and an RMS measurement. They are each what they are. To make proper use of these measurements you need to know the circuit and what is the goal of the measurements you are making. You get non-equal results in some equations when you use one method of measurement vs the other, and sometimes the same result occurs.
The question of this thread looks very much like it was a question on a test and Electric-Light wanted to see how members of this forum would respond.
What is your answer to the following question?
What is the maximum phase shift of an RC network?
.
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I answered ?A,? before I read any of the responses, and I got it wrong. The correct answer is ?C.?
There are two pieces of information that have the potential of making the answer something other than 1kW. We were given one of them, and not the other.
The equations Dennis gave us in post #2 presume that the power factor is 1.0. In this case, that is a valid presumption, since we were told (not in the text of post #1 but in the title of the thread) that we are dealing with an incandescent light. Those devices are purely resistive, so it is appropriate to infer that the power factor is 1.0.
The other piece of information that could cause power to be something other than 1kW is the nature of the power source. If it is a pure sine wave, then we can use Vrms x Irms = P. If it is not a pure sine wave, then all bets are off. So since we were not told the nature of the power source, that is a key piece of information that is missing. Therefore, the correct answer is ?C.?
There are two pieces of information that have the potential of making the answer something other than 1kW. We were given one of them, and not the other.
The equations Dennis gave us in post #2 presume that the power factor is 1.0. In this case, that is a valid presumption, since we were told (not in the text of post #1 but in the title of the thread) that we are dealing with an incandescent light. Those devices are purely resistive, so it is appropriate to infer that the power factor is 1.0.
The other piece of information that could cause power to be something other than 1kW is the nature of the power source. If it is a pure sine wave, then we can use Vrms x Irms = P. If it is not a pure sine wave, then all bets are off. So since we were not told the nature of the power source, that is a key piece of information that is missing. Therefore, the correct answer is ?C.?
I thought the very purpose of RMS was to take into account the difference between dc (which would very likely be a non-sinusoidal wave (flat, square, PWM etc.) ), ac, and non-sinusoidal waves, so that apples could be calculated with apples.... and that the purpose of stating the values as RMS would be so that the equation could be solved just that easily.
Do I think wrong?
Last edited:
gar
Senior Member
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realolman:
Yes you are wrong.
You can not just multiply Vrms and Irms and calculate load power from this. It may yield the correct result or it may not. You need more information. I discussed this in a previous post.
If you have a known pure stable resistance at the frequency of interest, then Irms^2 * R will give you the power. Same with voltage squared. But these fail if resistance is nonlinear.
For some circuits and waveforms you need to integrate the instantaneous product of voltage and current over some desired time and divide by that time period to get the average power over that period.
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realolman:
Yes you are wrong.
You can not just multiply Vrms and Irms and calculate load power from this. It may yield the correct result or it may not. You need more information. I discussed this in a previous post.
If you have a known pure stable resistance at the frequency of interest, then Irms^2 * R will give you the power. Same with voltage squared. But these fail if resistance is nonlinear.
For some circuits and waveforms you need to integrate the instantaneous product of voltage and current over some desired time and divide by that time period to get the average power over that period.
.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
So where is the Op on this. I hate when questions are asked and the OP disappears. I am curious what type of exam this question came from. I would be surprised if this was a question on an electrical contractor exam. I bet most of us don't know what RMS stands for never mind understanding the pure sine wave that Charlie talked about.
I am curious what type of exam this question came from. I would be surprised if this was a question on an electrical contractor exam. I bet most of us don't know what RMS stands for never mind understanding the pure sine wave that Charlie talked about. - Status